# To Vectory! – The Vector Racing Grand Prix

User CarpetPython posted a new take on this problem which puts a much bigger focus on heuristic solutions, due to an increased search space. I personally think that challenge is much nicer than mine, so consider giving that one a try!

Vector racing is an addictive game that can be played with a pen and a sheet of square-ruled paper. You draw an arbitrary racetrack on the paper, define a start and end and then you steer your point-sized car in a turn-based manner. Get to the end as fast as you can, but be careful not to end up in a wall!

## The Track

• The map is a two dimensional grid, where each cell has integer coordinates.
• You move on the grid cells.
• Each grid cell is either part of the track or it's a wall.
• Exactly one track cell is the starting coordinate.
• At least one track cell is designated as the goal. Landing on any of these completes the race. Multiple goal cells are not necessarily connected.

## Steering the Car

Your car starts at a given coordinate and with velocity vector (0, 0). In each turn, you may adjust each component of the velocity by ±1 or leave it as is. Then, the resulting velocity vector is added to your car's position.

A picture may help! The red circle was your position last turn. The blue circle is your current position. Your velocity is the vector from the red to the blue circle. In this turn, depending on how you adjust your velocity, you may move to any of the green circles.

If you land in a wall, you lose immediately.

You guessed it: write a program which, given a racetrack as input, steers the car to one of the goal cells in as few turns as possible. Your solution should be able to deal reasonably well with arbitrary tracks and not be specifically optimised towards the test cases provided.

### Input

target
n m
[ASCII representation of an n x m racetrack]
time


target is the maximum number of turns you may take to complete the track, and time is your total time budget for the track, in seconds (not necessarily integer). See below for details about timing.

For the newline-delimited track, the following characters are used:

• # – wall
• Sthe start
• *a goal
• . – all other track cells (i.e. road)

All cells outside the n x m grid provided are implied to be walls.

The coordinate origin is in the top left corner.

Here is a simple example:

8
4.0
9 6
###...***
###...***
###...***
......###
S.....###
......###


Using 0-based indexing, the starting coordinate would be (0,4).

After every move you will receive further input:

x y
u v
time


Where x, y, u, v are all 0-based integers. (x,y) is your current position and (u,v) is your current velocity. Note that x+u and/or y+v could be out of bounds.

time is whatever is left of your time budget, in seconds. Feel free to ignore this. This is only for participants who really want to take their implementation to the time limit.

When the game is over (because you've landed in a wall, moved out of bounds, exceeded target turns, ran out of time or reached the goal), you'll receive an empty line.

### Output

For each turn, write to stdout:

Δu Δv


where Δu and Δv each are one of -1, 0, 1. This will be added to (u,v) to determine your new position. Just to clarify, the directions are as follows

(-1,-1) ( 0,-1) ( 1,-1)
(-1, 0) ( 0, 0) ( 1, 0)
(-1, 1) ( 0, 1) ( 1, 1)


An optimal solution for the above example would be

1 0
1 -1
1 0


Note that the controller does not attach itself to stderr, so you're free to use it for any sort of debug output you may need while developing your bot. Please remove/comment out any such output in your submitted code, though.

Your bot may take half a second to respond in each single turn. For turns that take longer, you will have a time budget (per track) of target/2 seconds. Every time a turn takes longer than half a second, the additional time will be subtracted from your time budget. When your time budget hits zero the current race will be aborted.

New: For practical reasons, I have to set a memory limit (since memory seems to be more limiting than time for reasonable track sizes). Therefore, I will have to abort any test run where the racer uses more than 1GB of memory as measured by Process Explorer as Private Bytes.

### Scoring

There is a benchmark of 20 tracks. For each track:

• If you complete the track, your score is the number of moves you need to reach a goal cell divided by target.
• If you run out of time/memory or don't reach the goal before target turns have elapsed or you land in a wall/out of bounds at any time, your score is 2.

Your overall score is the sum of the individual track scores. Lowest score wins!

Furthermore, every participant may (and is strongly encouraged to) provide an additional benchmark track, which will be added to the official list. Previous answers will be re-evaluated including this new track. This is to make sure that no solution is tailored too closely to the existing test cases and to account for interesting edge cases I might have missed (but which you may spot).

### Tie breaking

Now that there is already an optimal solution, this will probably be the main factor for participants' scores.

If there is a tie (due to several answers solving all tracks optimally or otherwise), I will add additional (larger) test cases to break the tie. To avoid any human bias when creating those tie-breakers, these will be generated in a fixed manner:

• I will increase the side length n by 10 compared to the last track generated this way. (I may skip sizes if they don't break the tie.)
• The basis is this vector-graphic
• This will be rasterised at the desired resolution using this Mathematica snippet.
• The start is in the top left corner. In particular, it will be the left-most cell of the top-most row of that end of the track.
• The goal is in the bottom right corner. In particular, it will be the right-most cell of the bottom-most row of that end of the track.
• The target will 4*n.

The final track of the initial benchmark was already generated like this, with n = 50.

## The Controller

The program that tests the submissions is written in Ruby and can be found on GitHub along with the benchmark file I'll be using. There is also an example bot called randomracer.rb in there, which simply picks random moves. You can use its basic structure as a starting point for your bot to see how the communication works.

You can run your own bot against a track file of your choice as follows:

ruby controller.rb track_file_name command to run your racer


e.g.

ruby controller.rb benchmark.txt ruby randomracer.rb


The repository also contains two classes Point2D and Track. If your submission is written in Ruby, feel free to reuse those for your convenience.

Command-line switches

You can add command-line switches -v, -s, -t before the benchmark's file name. If you want to use multiple switches, you can also do, for instance, -vs. This is what they do:

-v (verbose): Use this to produce a little more debug output from the controller.

-s (silent): If you prefer to keep track of your position and velocity yourself and don't care about the time budget, you can switch off those three lines of output each turn (sent to your submission) with this flag.

-t (tracks): Lets you select individual tracks to be tested. E.g. -t "1,2,5..8,15" would test tracks 1, 2, 5, 6, 7, 8 and 15 only. Thanks a lot to Ventero for this feature and the options parser.

• If you're using randomness, please state this, so I can average your score over multiple runs.
• The code for your submission.
• The location of a free compiler or interpreter for your language of choice that runs on a Windows 8 machine.
• Compilation instructions if necessary.
• A Windows command-line string to run your submission.
• Whether your submission requires the -s flag or not.
• (optionally) A new, solvable track which will be added to the benchmark. I will determine a reasonable target for your track manually. When the track is added to the benchmark, I'll edit it out of your answer. I reserve the right to ask you for a different track (just in case you add an disproportionally large track, include obscene ASCII art in the track, etc.). When I add the test case to the benchmark set, I'll replace the track in your answer with a link to the track in the benchmark file to reduce the clutter in this post.

As you can see, I'll be testing all submissions on a Windows 8 machine. If there is absolutely no way to get your submission running on Windows, I can also try on a Ubuntu VM. This will be considerably slower though, so if you want to max out the time limit, make sure your program runs on Windows.

May the best driver emerge vectorious!

# But I wanna play!

If you want to try out the game yourself to get a better feel for it, there is this implementation. The rules used there are slightly more sophisticated, but it is similar enough to be useful, I think.

Last updated: 01/09/2014, 21:29 UTC
Tracks in benchmark: 25
Tie-breaker sizes: 290, 440

1. 6.86688 – Kuroi Neko
2. 8.73108 – user2357112 - 2nd submission
3. 9.86627 – nneonneo
4. 10.66109 – user2357112 - 1st submission
5. 12.49643 – Ray
6. 40.0759 – pseudonym117 (probabilistic)

Detailed test results. (The scores for probabilistic submissions have been determined separately.)

# C++11 - 6.66109

Yet another breadth first search implementation, only optimized.

It must be run with the -s option.
Its input is not sanitized at all, so wrong tracks might turn it into a pumpkin.

I tested it with Microsoft Visual C++ 2013, release build with the default /O2 flag (optimize for speed).
DOES build OK with g++ and the Microsoft IDE.
My barebone memory allocator is a piece of crap, so don't expect it to work with other STL implementations of unordered_set!

#include <cstdint>
#include <iostream>
#include <fstream>
#include <sstream>
#include <queue>
#include <unordered_set>

#define MAP_START 'S'
#define MAP_WALL  '#'
#define MAP_GOAL  '*'

#define NODE_CHUNK_SIZE   100 // increasing this will not improve performances
#define VISIT_CHUNK_SIZE 1024 // increasing this will slightly reduce mem consumption at the (slight) cost of speed

#define HASH_POS_BITS 8 // number of bits for one coordinate
#define HASH_SPD_BITS (sizeof(size_t)*8/2-HASH_POS_BITS)

typedef int32_t tCoord; // 32 bits required to overcome the 100.000 cells (insanely) long challenge

// basic vector arithmetics
struct tPoint {
tCoord x, y;
tPoint(tCoord x = 0, tCoord y = 0) : x(x), y(y) {}
tPoint operator+ (const tPoint & p) { return tPoint(x + p.x, y + p.y); }
tPoint operator- (const tPoint & p) { return tPoint(x - p.x, y - p.y); }
bool operator== (const tPoint & p) const { return p.x == x && p.y == y;  }
};

// a barebone block allocator. Improves speed by about 30%
template <class T, size_t SIZE> class tAllocator
{
T * chunk;
size_t i_alloc;
size_t m_alloc;
public:
typedef T                 value_type;
typedef value_type*       pointer;
typedef const value_type* const_pointer;
typedef std::size_t       size_type;
typedef value_type&       reference;
typedef const value_type& const_reference;
tAllocator()                                              { m_alloc = i_alloc = SIZE; }
template <class U> tAllocator(const tAllocator<U, SIZE>&) { m_alloc = i_alloc = SIZE; }
template <class U> struct rebind { typedef tAllocator<U, SIZE> other; };
pointer allocate(size_type n, const_pointer = 0)
{
if (n > m_alloc) { i_alloc = m_alloc = n; }      // grow max size if request exceeds capacity
if ((i_alloc + n) > m_alloc) i_alloc = m_alloc;  // dump current chunk if not enough room available
if (i_alloc == m_alloc) { chunk = new T[m_alloc]; i_alloc = 0; } // allocate new chunk when needed
T * mem = &chunk[i_alloc];
i_alloc += n;
return mem;
}
void deallocate(pointer, size_type) { /* memory is NOT released until process exits */ }
void construct(pointer p, const value_type& x) { new(p)value_type(x); }
void destroy(pointer p) { p->~value_type(); }
};

// a node in our search graph
class tNode {
static tAllocator<tNode, NODE_CHUNK_SIZE> mem; // about 10% speed gain over a basic allocation
tNode * parent;
public:
tPoint pos;
tPoint speed;
static tNode * alloc (tPoint pos, tPoint speed, tNode * parent) { return new (mem.allocate(1)) tNode(pos, speed, parent); }
tNode (tPoint pos = tPoint(), tPoint speed = tPoint(), tNode * parent = nullptr) : parent(parent), pos(pos), speed(speed) {}
bool operator== (const tNode& n) const { return n.pos == pos && n.speed == speed; }
void output(void)
{
std::string output;
tPoint v = this->speed;
for (tNode * n = this->parent ; n != nullptr ; n = n->parent)
{
tPoint a = v - n->speed;
v = n->speed;
std::ostringstream ss;  // a bit of shitty c++ text I/O to print elements in reverse order
ss << a.x << ' ' << a.y << '\n';
output = ss.str() + output;
}
std::cout << output;
}
};
tAllocator<tNode, NODE_CHUNK_SIZE> tNode::mem;

// node queueing and storing
static int num_nodes = 0;
class tNodeJanitor {
// set of already visited nodes. Block allocator improves speed by about 20%
struct Hasher { size_t operator() (tNode * const n) const
{
int64_t hash = // efficient hashing is the key of performances
((int64_t)n->pos.x   << (0 * HASH_POS_BITS))
^ ((int64_t)n->pos.y   << (1 * HASH_POS_BITS))
^ ((int64_t)n->speed.x << (2 * HASH_POS_BITS + 0 * HASH_SPD_BITS))
^ ((int64_t)n->speed.y << (2 * HASH_POS_BITS + 1 * HASH_SPD_BITS));
return (size_t)((hash >> 32) ^ hash);
//return (size_t)(hash);
}
};
struct Equalizer { bool operator() (tNode * const n1, tNode * const n2) const
{ return *n1 == *n2; }};
std::unordered_set<tNode *, Hasher, Equalizer, tAllocator<tNode *, VISIT_CHUNK_SIZE>> visited;
std::queue<tNode *> queue; // currently explored nodes queue
public:
bool empty(void) { return queue.empty();  }
tNode * dequeue() { tNode * n = queue.front(); queue.pop(); return n; }
tNode * enqueue_if_new (tPoint pos, tPoint speed = tPoint(0,0), tNode * parent = nullptr)
{
tNode signature (pos, speed);
tNode * n = nullptr;
if (visited.find (&signature) == visited.end()) // the classy way to check if an element is in a set
{
n = tNode::alloc(pos, speed, parent);
queue.push(n);
visited.insert (n);
num_nodes++;
}
return n;
}
};

// map representation
class tMap {
std::vector<char> cell;
tPoint dim; // dimensions
public:
void set_size(tCoord x, tCoord y) { dim = tPoint(x, y); cell.resize(x*y); }
void set(tCoord x, tCoord y, char c) { cell[y*dim.x + x] = c; }
char get(tPoint pos)
{
if (pos.x < 0 || pos.x >= dim.x || pos.y < 0 || pos.y >= dim.y) return MAP_WALL;
return cell[pos.y*dim.x + pos.x];
}
void dump(void)
{
for (int y = 0; y != dim.y; y++)
{
for (int x = 0; x != dim.x; x++) fprintf(stderr, "%c", cell[y*dim.x + x]);
fprintf(stderr, "\n");
}
}
};

// race manager
class tRace {
tPoint start;
tNodeJanitor border;
static tPoint acceleration[9];
public:
tMap map;
tRace ()
{
int target;
tCoord sx, sy;
std::cin >> target >> sx >> sy;
std::cin.ignore();
map.set_size (sx, sy);
std::string row;
for (int y = 0; y != sy; y++)
{
std::getline(std::cin, row);
for (int x = 0; x != sx; x++)
{
char c = row[x];
if (c == MAP_START) start = tPoint(x, y);
map.set(x, y, c);
}
}
}

// all the C++ crap above makes for a nice and compact solver
tNode * solve(void)
{
tNode * initial = border.enqueue_if_new (start);
while (!border.empty())
{
tNode * node = border.dequeue();
tPoint p = node->pos;
tPoint v = node->speed;
for (tPoint a : acceleration)
{
tPoint nv = v + a;
tPoint np = p + nv;
char c = map.get(np);
if (c == MAP_WALL) continue;
if (c == MAP_GOAL) return new tNode (np, nv, node);
border.enqueue_if_new (np, nv, node);
}
}
return initial; // no solution found, will output nothing
}
};
tPoint tRace::acceleration[] = {
tPoint(-1,-1), tPoint(-1, 0), tPoint(-1, 1),
tPoint( 0,-1), tPoint( 0, 0), tPoint( 0, 1),
tPoint( 1,-1), tPoint( 1, 0), tPoint( 1, 1)};

#include <ctime>
int main(void)
{
tRace race;
clock_t start = clock();
tNode * solution = race.solve();
std::cerr << "time: " << (clock()-start)/(CLOCKS_PER_SEC/1000) << "ms nodes: " << num_nodes << std::endl;
solution->output();
return 0;
}


# Results

 No.       Size     Target   Score     Details
-------------------------------------------------------------------------------------
1       37 x 1        36   0.22222   Racer reached goal at ( 36, 0) in 8 turns.
2       38 x 1        37   0.24324   Racer reached goal at ( 37, 0) in 9 turns.
3       33 x 1        32   0.25000   Racer reached goal at ( 32, 0) in 8 turns.
4       10 x 10       10   0.40000   Racer reached goal at ( 7, 7) in 4 turns.
5        9 x 6         8   0.37500   Racer reached goal at ( 6, 0) in 3 turns.
6       15 x 7        16   0.37500   Racer reached goal at ( 12, 4) in 6 turns.
7       17 x 8        16   0.31250   Racer reached goal at ( 14, 0) in 5 turns.
8       19 x 13       18   0.27778   Racer reached goal at ( 0, 11) in 5 turns.
9       60 x 10      107   0.14953   Racer reached goal at ( 0, 6) in 16 turns.
10       31 x 31      106   0.23585   Racer reached goal at ( 27, 0) in 25 turns.
11       31 x 31      106   0.24528   Racer reached goal at ( 15, 15) in 26 turns.
12       50 x 20       50   0.24000   Racer reached goal at ( 49, 10) in 12 turns.
13      100 x 100    2600   0.01385   Racer reached goal at ( 50, 0) in 36 turns.
14       79 x 63      242   0.24380   Racer reached goal at ( 3, 42) in 59 turns.
15       26 x 1        25   0.32000   Racer reached goal at ( 25, 0) in 8 turns.
16       17 x 1        19   0.52632   Racer reached goal at ( 16, 0) in 10 turns.
17       50 x 1        55   0.34545   Racer reached goal at ( 23, 0) in 19 turns.
18       10 x 7        23   0.34783   Racer reached goal at ( 1, 3) in 8 turns.
19       55 x 55       45   0.17778   Racer reached goal at ( 50, 26) in 8 turns.
20      101 x 100     100   0.14000   Racer reached goal at ( 99, 99) in 14 turns.
21   100000 x 1         1   1.00000   Racer reached goal at ( 0, 0) in 1 turns.
22       50 x 50      200   0.05500   Racer reached goal at ( 47, 46) in 11 turns.
23      290 x 290    1160   0.16466   Racer reached goal at ( 269, 265) in 191 turns.
-------------------------------------------------------------------------------------
TOTAL SCORE:                 6.66109


# Performances

That crappy C++ language has a knack to make you jump through hoops just to move a matchstick. However, you can whip it into producing relatively fast and memory-efficient code.

## Hashing

Here the key is to provide a good hash table for the nodes. That is by far the dominant factor for execution speed.
Two implementations of unordered_set (GNU and Microsoft) yielded a 30% execution speed difference (in favor of GNU, yay!).

The difference is not really surprising, what with the truckloads of code hidden behind unordered_set.

Out of curiosity, I did some stats on the final state of the hash table.
Both algorithms end up with nearly the same bucket/elements ratio, but repartition varies:
for the 290x290 tie breaker, GNU gets an average of 1.5 elements per non empty bucket, while Microsoft is at 5.8 (!).

Looks like my hashing function is not very well randomized by Microsoft... I wonder if the guys in Redmond did really benchmark their STL, or maybe my use case just favors the GNU implementation...

Sure, my hashing function is nowhere near optimal. I could have used the usual integer mixing based on multiple shift/muls but a hash-efficient function takes time to compute.

It appears the number of hash table queries is very high compared with the number of insertions. For instance, in the 290x290 tie breaker, you have about 3.6 million insertions for 22.7 million queries.
In this context, a suboptimal but fast hashing yields better performances.

## Memory allocation

Providing an efficient memory allocator comes second. It improved performances by about 30%. Whether it is worth the added crap code is debatable :).

The current version uses between 40 and 55 bytes per node.
The functional data require 24 bytes for a node (4 coordinates and 2 pointers).
Due to the insane 100.000 lines test case, coordinates have to be stored in 4 bytes words, otherwise you could gain 8 bytes by using shorts (with a max coordinate value of 32767). The remaining bytes are mostly consumed by the unordered set's hash table. It means the data handling actually consumes a bit more than the "useful" payload.

## And the winner is...

On my PC under Win7, the tie breaker (case 23, 290x290) is solved by the worst version (i.e. Microsoft compiled) in about 2.2 seconds, with a memory consumption of about 185 Mb.
For comparison, the current leader (python code by user2357112) takes a bit more than 30 seconds and consumes about 780 Mb.

# Controller issues

I'm not quite sure I might be able to code in Ruby to save my life.
However, I spotted and hacked two problems out of the controller code:

## 1) map reading in track.rb

With ruby 1.9.3 installed, the track reader would croak about shift.to_i not being available for string.lines.
After a long bit of wading through the online Ruby documentation I gave up on strings and used an intermediate array instead, like so (right at the begining of the file):

def initialize(string)
@track = Array.new;
string.lines.each do |line|
@track.push (line.chomp)
end


## 2) killing ghosts in controller.rb

As other posters have already noted, the controller sometimes tries to kill processes that have already exited. To avoid these disgracious error outputs, I simply handled the exception away, like so (around line 134):

if silent
begin # ;!;
Process.kill('KILL', racer.pid)
rescue Exception => e
end


# Test case

To defeat the brute force approach of the BFS solvers, the worst track is the opposite of the 100.000 cells map: a completely free area with the goal as far from the start as possible.

In this example, a 100x400 map with the goal in the upper left corner and the start in the lower right.

This map has a solution in 28 turns, but a BFS solver will explore millions of states to find it (mine counted 10.022.658 states visited, took about 12 seconds and peaked at 600 Mb !).

With less than half the surface of the 290x290 tie breaker, it requires about 3 times more node visits. On the other hand, a heuristic / A* based solver should defeat it easily.

30
100 400
*...................................................................................................
....................................................................................................
< 400 lines in all >
....................................................................................................
....................................................................................................
...................................................................................................S


# Bonus: an equivalent (but somewhat less efficient) PHP version

This is what I started with, before the inherent language slowness convinced me to use C++.
PHP internal hash tables don't seem as efficient as Python's, at least in this particular case :).

<?php

class Trace {
static $file; public static$state_member;
public static $state_target; static function msg ($msg)
{
fputs (self::$file, "$msg\n");
}

static function dump ($var,$msg=null)
{
ob_start();
if ($msg) echo "$msg ";
var_dump($var);$dump=ob_get_contents();
ob_end_clean();
fputs (self::$file, "$dump\n");
}

function init ($fname) { self::$file = fopen ($fname, "w"); } } Trace::init ("racer.txt"); class Point { public$x;
public $y; function __construct ($x=0, $y=0) {$this->x = (float)$x;$this->y = (float)$y; } function __toString () { return "[$this->x $this->y]"; } function add ($v)
{
return new Point ($this->x +$v->x, $this->y +$v->y);
}

function vector_to ($goal) { return new Point ($goal->x - $this->x,$goal->y - $this->y); } } class Node { public$posx  , $posy ; public$speedx, $speedy; private$parent;

public function __construct ($posx,$posy, $speedx,$speedy, $parent) {$this->posx = $posx;$this->posy = $posy;$this->speedx = $speedx;$this->speedy = $speedy;$this->parent = $parent; } public function path () {$res = array();
$v = new Point ($this->speedx, $this->speedy); for ($node = $this->parent ;$node != null ; $node =$node->parent)
{
$nv = new Point ($node->speedx, $node->speedy);$a = $nv->vector_to ($v);
$v = new Point ($node->speedx, $node->speedy); array_unshift ($res, $a); } return$res;
}
}

class Map {

private static $target; // maximal number of turns private static$time;         // time available to solve
private static $sx,$sy;      // map dimensions
private static $cell; // cells of the map private static$start;        // starting point
private static $acceleration; // possible acceleration values public static function init () { // read map definition self::$target = trim(fgets(STDIN));
list (self::$sx, self::$sy) = explode (" ", trim(fgets(STDIN)));
self::$cell = array(); for ($y = 0 ; $y != self::$sy ; $y++) self::$cell[] = str_split (trim(fgets(STDIN)));
self::$time = trim(fgets(STDIN)); // get starting point foreach (self::$cell as $y=>$row)
{
$x = array_search ("S",$row);
if ($x !== false) { self::$start = new Point ($x,$y);
Trace::msg ("start ".self::$start); break; } } // compute possible acceleration values self::$acceleration = array();
for ($x = -1 ;$x <= 1 ; $x++) for ($y = -1 ; $y <= 1 ;$y++)
{
self::$acceleration[] = new Point ($x, $y); } } public static function solve () {$now = microtime(true);
$res = array();$border = array (new Node (self::$start->x, self::$start->y, 0, 0, null));
$present = array (self::$start->x." ".self::$start->y." 0 0" => 1); while (count ($border))
{
if ((microtime(true) - $now) > 1) { Trace::msg (count($present)." nodes, ".round(memory_get_usage(true)/1024)."K");
$now = microtime(true); }$node = array_shift ($border); //Trace::msg ("node$node->pos $node->speed");$px = $node->posx;$py = $node->posy;$vx = $node->speedx;$vy = $node->speedy; foreach (self::$acceleration as $a) {$nvx = $vx +$a->x;
$nvy =$vy + $a->y;$npx = $px +$nvx;
$npy =$py + $nvy; if ($npx < 0 || $npx >= self::$sx || $npy < 0 ||$npy >= self::$sy || self::$cell[$npy][$npx] == "#")
{
//Trace::msg ("invalid position $px,$py $vx,$vy -> $npx,$npy");
continue;
}
if (self::$cell[$npy][$npx] == "*") { Trace::msg ("winning position$px,$py$vx,$vy ->$npx,$npy");$end = new Node ($npx,$npy, $nvx,$nvy, $node);$res = $end->path (); break 2; } //Trace::msg ("checking$np $nv");$signature = "$npx$npy $nvx$nvy";
if (isset ($present[$signature])) continue;
//Trace::msg ("*** adding $np$nv");
$border[] = new Node ($npx, $npy,$nvx, $nvy,$node);
$present[$signature] = 1;
}
}
return $res; } } ini_set("memory_limit","1000M"); Map::init ();$res = Map::solve();
//Trace::dump ($res); foreach ($res as $a) echo "$a->x $a->y\n"; ?>  • erf... My barebone allocator is just a bit too barebone. I'll add the necessary crap to make it work with g++, then. Sorry about that. – user16991 Aug 31 '14 at 13:23 • OK, that's fixed. The g++ version even actually works about 30% faster. It now outputs some stats on stderr. Feel free to comment it out (from the last lines of the source). Sorry again for the blunder. – user16991 Aug 31 '14 at 16:11 • Alright, it works now and I reproduced your score. It's damn fast! :) I'll add your test case to the benchmark, but I'll change the target to 400, as that's in line with how I determined all the other targets (except the tie breaker). I'll update the main post once I've reran all the other submissions. – Martin Ender Aug 31 '14 at 20:13 • Updated the results. There was no need for a tie breaker, because all other submissions exceed the memory limit on your test track. Congrats on that! :) – Martin Ender Aug 31 '14 at 20:32 • Thanks. Actually this challenge gave me an occasion to dig into these STL hash tables. Though I hate C++ guts, I can't help but getting killed by my curiosity. Meow! :). – user16991 Aug 31 '14 at 20:41 # C++, 5.4 (deterministic, optimal) Dynamic programming solution. Provably optimal. Very fast: solves all 20 testcases in 0.2s. Should be especially fast on 64-bit machines. Assumes the board is less than 32,000 places in each direction (which should hopefully be true). This racer is a little unusual. It computes the optimal path at the starting line, and afterwards executes the computed path instantly. It ignores the time control and assumes it can finish the optimization step on time (which should be true for any reasonably modern hardware). On excessively large maps, there's a small chance that the racer may segfault. If you can convince it to segfault, you get a brownie point, and I'll fix it to use an explicit loop. Compile with g++ -O3. May require C++11 (for <unordered_map>). To run, just run the compiled executable (no flags or options are supported; all input is taken on stdin). #include <unordered_map> #include <iostream> #include <string> #include <vector> #include <sstream> #include <cstdint> #define MOVES_INF (1<<30) union state { struct { short px, py, vx, vy; }; uint64_t val; }; struct result { int nmoves; short dvx, dvy; }; typedef std::unordered_map<uint64_t, result> cache_t; int target, n, m; std::vector<std::string> track; cache_t cache; static int solve(uint64_t val) { cache_t::iterator it = cache.find(val); if(it != cache.end()) return it->second.nmoves; // prevent recursion result res; res.nmoves = MOVES_INF; cache[val] = res; state cur; cur.val = val; for(int dvx = -1; dvx <= 1; dvx++) for(int dvy = -1; dvy <= 1; dvy++) { state next; next.vx = cur.vx + dvx; next.vy = cur.vy + dvy; next.px = cur.px + next.vx; next.py = cur.py + next.vy; if(next.px < 0 || next.px >= n || next.py < 0 || next.py >= m) continue; char c = track[next.py][next.px]; if(c == '*') { res.nmoves = 1; res.dvx = dvx; res.dvy = dvy; break; } else if(c == '#') { continue; } else { int score = solve(next.val) + 1; if(score < res.nmoves) { res.nmoves = score; res.dvx = dvx; res.dvy = dvy; } } } cache[val] = res; return res.nmoves; } bool solve_one() { std::string line; float time; std::cin >> target; // std::cin >> time; // uncomment to use "time" control std::cin >> n >> m; if(!std::cin) return false; std::cin.ignore(); // skip newline at end of "n m" line track.clear(); track.reserve(m); for(int i=0; i<m; i++) { std::getline(std::cin, line); track.push_back(line); } cache.clear(); state cur; cur.vx = cur.vy = 0; for(int y=0; y<m; y++) for(int x=0; x<n; x++) { if(track[y][x] == 'S') { cur.px = x; cur.py = y; break; } } solve(cur.val); int sol_len = 0; while(track[cur.py][cur.px] != '*') { cache_t::iterator it = cache.find(cur.val); if(it == cache.end() || it->second.nmoves >= MOVES_INF) { std::cerr << "Failed to solve at p=" << cur.px << "," << cur.py << " v=" << cur.vx << "," << cur.vy << std::endl; return true; } int dvx = it->second.dvx; int dvy = it->second.dvy; cur.vx += dvx; cur.vy += dvy; cur.px += cur.vx; cur.py += cur.vy; std::cout << dvx << " " << dvy << std::endl; sol_len++; } //std::cerr << "Score: " << ((float)sol_len) / target << std::endl; return true; } int main() { /* benchmarking: */ //while(solve_one()) // ; /* regular running */ solve_one(); std::string line; while(std::cin) std::getline(std::cin, line); return 0; }  ## Results  No. Size Target Score Details ------------------------------------------------------------------------------------- 1 37 x 1 36 0.22222 Racer reached goal at ( 36, 0) in 8 turns. 2 38 x 1 37 0.24324 Racer reached goal at ( 37, 0) in 9 turns. 3 33 x 1 32 0.25000 Racer reached goal at ( 32, 0) in 8 turns. 4 10 x 10 10 0.40000 Racer reached goal at ( 7, 7) in 4 turns. 5 9 x 6 8 0.37500 Racer reached goal at ( 6, 0) in 3 turns. 6 15 x 7 16 0.37500 Racer reached goal at ( 12, 4) in 6 turns. 7 17 x 8 16 0.31250 Racer reached goal at ( 15, 0) in 5 turns. 8 19 x 13 18 0.27778 Racer reached goal at ( 1, 11) in 5 turns. 9 60 x 10 107 0.14953 Racer reached goal at ( 2, 6) in 16 turns. 10 31 x 31 106 0.25472 Racer reached goal at ( 28, 0) in 27 turns. 11 31 x 31 106 0.24528 Racer reached goal at ( 15, 15) in 26 turns. 12 50 x 20 50 0.24000 Racer reached goal at ( 49, 10) in 12 turns. 13 100 x 100 2600 0.01385 Racer reached goal at ( 50, 0) in 36 turns. 14 79 x 63 242 0.26860 Racer reached goal at ( 3, 42) in 65 turns. 15 26 x 1 25 0.32000 Racer reached goal at ( 25, 0) in 8 turns. 16 17 x 1 19 0.52632 Racer reached goal at ( 16, 0) in 10 turns. 17 50 x 1 55 0.34545 Racer reached goal at ( 23, 0) in 19 turns. 18 10 x 7 23 0.34783 Racer reached goal at ( 1, 3) in 8 turns. 19 55 x 55 45 0.17778 Racer reached goal at ( 52, 26) in 8 turns. 20 50 x 50 200 0.05500 Racer reached goal at ( 47, 46) in 11 turns. ------------------------------------------------------------------------------------- TOTAL SCORE: 5.40009  ## New Testcase • Well, something like this was pretty much expected. There just isn't enough state in the puzzle to render dynamic programming infeasible. If I enter, I'm going to need to submit a map that requires more sophisticated search strategies to solve. – user2357112 supports Monica Jun 27 '14 at 1:27 • How does your racer perform on your test case? – user2357112 supports Monica Jun 27 '14 at 1:28 • 0.14 (14 moves) – nneonneo Jun 27 '14 at 1:46 • Is that time taken or moves/target? If it's moves/target, how does it perform in terms of time? – user2357112 supports Monica Jun 27 '14 at 1:50 • I think I've found a bug in the cycle prevention code. It assumes that for each state the search reaches from a state S, an optimal path cannot be to return to S. It might seem that if an optimal path does return to S, then the state cannot be on an optimal path (since we could just remove the loop it's on and get a shorter path), and so we don't care whether we get a too-high result for that state. However, if an optimal path passes through states A and B in that order, but the search first finds A while B is still on the stack, then A's results are poisoned by the loop prevention. – user2357112 supports Monica Jun 28 '14 at 21:43 ## Python 2, deterministic, optimal Here's my racer. I haven't tested it on the benchmark (still waffling about what version and installer of Ruby to install), but it should solve everything optimally and under the time limit. The command to run it is python whateveryoucallthefile.py. Needs the -s controller flag. # Breadth-first search. # Future directions: bidirectional search and/or A*. import operator import time acceleration_options = [(dvx, dvy) for dvx in [-1, 0, 1] for dvy in [-1, 0, 1]] class ImpossibleRaceError(Exception): pass def read_input(line_source=raw_input): # We don't use the target. target = int(line_source()) width, height = map(int, line_source().split()) input_grid = [line_source() for _ in xrange(height)] start = None for i in xrange(height): for j in xrange(width): if input_grid[i][j] == 'S': start = i, j break if start is not None: break walls = [[cell == '#' for cell in row] for row in input_grid] goals = [[cell == '*' for cell in row] for row in input_grid] return start, walls, goals def default_bfs_stop_threshold(walls, goals): num_not_wall = sum(sum(map(operator.not_, row)) for row in walls) num_goals = sum(sum(row) for row in goals) return num_goals * num_not_wall def bfs(start, walls, goals, stop_threshold=None): if stop_threshold is None: stop_threshold = default_bfs_stop_threshold(walls, goals) # State representation is (x, y, vx, vy) x, y = start initial_state = (x, y, 0, 0) frontier = {initial_state} # Visited set is tracked by a map from each state to the last move taken # before reaching that state. visited = {initial_state: None} while len(frontier) < stop_threshold: if not frontier: raise ImpossibleRaceError new_frontier = set() for x, y, vx, vy in frontier: for dvx, dvy in acceleration_options: new_vx, new_vy = vx+dvx, vy+dvy new_x, new_y = x+new_vx, y+new_vy new_state = (new_x, new_y, new_vx, new_vy) if not (0 <= new_x < len(walls) and 0 <= new_y < len(walls[0])): continue if walls[new_x][new_y]: continue if new_state in visited: continue new_frontier.add(new_state) visited[new_state] = dvx, dvy if goals[new_x][new_y]: return construct_path_from_bfs(new_state, visited) frontier = new_frontier def construct_path_from_bfs(goal_state, best_moves): reversed_path = [] current_state = goal_state while best_moves[current_state] is not None: move = best_moves[current_state] reversed_path.append(move) x, y, vx, vy = current_state dvx, dvy = move old_x, old_y = x-vx, y-vy # not old_vx or old_vy old_vx, old_vy = vx-dvx, vy-dvy current_state = (old_x, old_y, old_vx, old_vy) return reversed_path[::-1] def main(): t = time.time() start, walls, goals = read_input() path = bfs(start, walls, goals, float('inf')) for dvx, dvy in path: # I wrote the whole program with x pointing down and y pointing right. # Whoops. Gotta flip things for the output. print dvy, dvx if __name__ == '__main__': main()  After inspecting nneonneo's racer (but not actually testing it, since I also don't have a C++ compiler), I have found that it seems to perform a nearly-exhaustive search of the state space, no matter how close the goal or how short a path has already been found. I have also found that the time rules mean building a map with a long, complex solution requires a long, boring time limit. Thus, my map submission is pretty simple: ## New Testcase (GitHub cannot display the long line. The track is *S.......[and so on].....) ### Additional submission: Python 2, bidirectional search This is an approach I wrote up about two months ago, when trying to optimize my breadth-first submission. For the test cases that existed at the time, it didn't offer an improvement, so I didn't submit it, but for kuroi's new map, it seems to just barely squeeze by under the memory cap. I'm still expecting kuroi's solver to beat this, but I'm interested in how it holds up. # Bidirectional search. # Future directions: A*. import operator import time acceleration_options = [(dvx, dvy) for dvx in [-1, 0, 1] for dvy in [-1, 0, 1]] class ImpossibleRaceError(Exception): pass def read_input(line_source=raw_input): # We don't use the target. target = int(line_source()) width, height = map(int, line_source().split()) input_grid = [line_source() for _ in xrange(height)] start = None for i in xrange(height): for j in xrange(width): if input_grid[i][j] == 'S': start = i, j break if start is not None: break walls = [[cell == '#' for cell in row] for row in input_grid] goals = [[cell == '*' for cell in row] for row in input_grid] return start, walls, goals def bfs_to_bidi_threshold(walls, goals): num_not_wall = sum(sum(map(operator.not_, row)) for row in walls) num_goals = sum(sum(row) for row in goals) return num_goals * (num_not_wall - num_goals) class GridBasedGoalContainer(object): '''Supports testing whether a state is a goal state with in. Does not perform bounds checking.''' def __init__(self, goal_grid): self.goal_grid = goal_grid def __contains__(self, state): x, y, vx, vy = state return self.goal_grid[x][y] def forward_step(state, acceleration): x, y, vx, vy = state dvx, dvy = acceleration new_vx, new_vy = vx+dvx, vy+dvy new_x, new_y = x+new_vx, y+new_vy return (new_x, new_y, new_vx, new_vy) def backward_step(state, acceleration): x, y, vx, vy = state dvx, dvy = acceleration old_x, old_y = x-vx, y-vy old_vx, old_vy = vx-dvx, vy-dvy return (old_x, old_y, old_vx, old_vy) def bfs(start, walls, goals): x, y = start initial_state = (x, y, 0, 0) initial_frontier = {initial_state} visited = {initial_state: None} goal_state, frontier, visited = general_bfs( frontier=initial_frontier, visited=visited, walls=walls, goalcontainer=GridBasedGoalContainer(goals), stop_threshold=float('inf'), step_function=forward_step ) return construct_path_from_bfs(goal_state, visited) def general_bfs( frontier, visited, walls, goalcontainer, stop_threshold, step_function): while len(frontier) <= stop_threshold: if not frontier: raise ImpossibleRaceError new_frontier = set() for state in frontier: for accel in acceleration_options: new_state = new_x, new_y, new_vx, new_vy = \ step_function(state, accel) if not (0 <= new_x < len(walls) and 0 <= new_y < len(walls[0])): continue if walls[new_x][new_y]: continue if new_state in visited: continue new_frontier.add(new_state) visited[new_state] = accel if new_state in goalcontainer: return new_state, frontier, visited frontier = new_frontier return None, frontier, visited def max_velocity_component(n): # It takes a distance of at least 0.5*v*(v+1) to achieve a velocity of # v in the x or y direction. That means the map has to be at least # 1 + 0.5*v*(v+1) rows or columns long to accomodate such a velocity. # Solving for v, we get a velocity cap as follows. return int((2*n-1.75)**0.5 - 0.5) def solver( start, walls, goals, mode='bidi'): x, y = start initial_state = (x, y, 0, 0) initial_frontier = {initial_state} visited = {initial_state: None} if mode == 'bidi': stop_threshold = bfs_to_bidi_threshold(walls, goals) elif mode == 'bfs': stop_threshold = float('inf') else: raise ValueError('Unsupported search mode: {}'.format(mode)) goal_state, frontier, visited = general_bfs( frontier=initial_frontier, visited=visited, walls=walls, goalcontainer=GridBasedGoalContainer(goals), stop_threshold=stop_threshold, step_function=forward_step ) if goal_state is not None: return construct_path_from_bfs(goal_state, visited) # Switching to bidirectional search. not_walls_or_goals = [] goal_list = [] for x in xrange(len(walls)): for y in xrange(len(walls[0])): if not walls[x][y] and not goals[x][y]: not_walls_or_goals.append((x, y)) if goals[x][y]: goal_list.append((x, y)) max_vx = max_velocity_component(len(walls)) max_vy = max_velocity_component(len(walls[0])) reverse_visited = {(goal_x, goal_y, goal_x-prev_x, goal_y-prev_y): None for goal_x, goal_y in goal_list for prev_x, prev_y in not_walls_or_goals if abs(goal_x-prev_x) <= max_vx and abs(goal_y - prev_y) <= max_vy} reverse_frontier = set(reverse_visited) while goal_state is None: goal_state, reverse_frontier, reverse_visited = general_bfs( frontier=reverse_frontier, visited=reverse_visited, walls=walls, goalcontainer=frontier, stop_threshold=len(frontier), step_function=backward_step ) if goal_state is not None: break goal_state, frontier, visited = general_bfs( frontier=frontier, visited=visited, walls=walls, goalcontainer=reverse_frontier, stop_threshold=len(reverse_frontier), step_function=forward_step ) forward_path = construct_path_from_bfs(goal_state, visited) backward_path = construct_path_from_bfs(goal_state, reverse_visited, step_function=forward_step) return forward_path + backward_path[::-1] def construct_path_from_bfs(goal_state, best_moves, step_function=backward_step): reversed_path = [] current_state = goal_state while best_moves[current_state] is not None: move = best_moves[current_state] reversed_path.append(move) current_state = step_function(current_state, move) return reversed_path[::-1] def main(): start, walls, goals = read_input() t = time.time() path = solver(start, walls, goals) for dvx, dvy in path: # I wrote the whole program with x pointing down and y pointing right. # Whoops. Gotta flip things for the output. print dvy, dvx if __name__ == '__main__': main()  • This sometimes fail on case 12 and 13. Dont know why since the error messages are somewhat ... unfriendly – Ray Jul 1 '14 at 8:20 • @Ray I do get error messages, too, but I always get the result for those anyway. I think it might rather be something in my controller, because it looks like the controller tries to kill the racer process although it's already finished. – Martin Ender Jul 1 '14 at 9:10 • @m.buettner I found the reason, add -s then it will be OK. – Ray Jul 1 '14 at 10:56 • @Ray Oh yeah, I'm doing that. I still get an error on tracks 13 and 14 when the controller is trying to kill the process although the result is already there. I guess I should look into that, but doesn't affect the scoring so I didn't bother yet. – Martin Ender Jul 1 '14 at 10:57 • Unfortunately, I had to add another rule. Memory seems to be more limiting than time in this challenge, so I had to set a hard to limit on memory consumption. Any run in which your racer uses more than 1GB of memory will be aborted to the same effect as exceed the time limit. For the current set of tracks, your score has not been affected by this change. (I think you'll reach that limit on tie-breakers around n = 400.) Please let me know if you apply any optimisations, so I can rerun the tests. – Martin Ender Jul 1 '14 at 14:42 # Python 3: 6.49643 (Optimal, BFS) For the old 20 case benchmark file, it got a score of 5.35643. The solution by @nneonneo is not optimal since it got 5.4. Some bugs maybe. This solution use BFS to search the Graph, each search state is in the form of (x, y, dx, dy). Then I use a map to map from states to distances. In the worst case, it's time and space complexity is O(n^2 m^2). This will rarely happen since the speed won't be too high or the racer will crash. Actually, it cost 3 seconds on my machine to complete all the 22 testcases. from collections import namedtuple, deque import itertools Field = namedtuple('Map', 'n m grids') class Grid: WALL = '#' EMPTY = '.' START = 'S' END = '*' def read_nums(): return list(map(int, input().split())) def read_field(): m, n = read_nums() return Field(n, m, [input() for i in range(n)]) def find_start_pos(field): return next((i, j) for i in range(field.n) for j in range(field.m) if field.grids[i][j] == Grid.START) def can_go(field, i, j): return 0 <= i < field.n and 0 <= j < field.m and field.grids[i][j] != Grid.WALL def trace_path(start, end, prev): if end == start: return end, step = prev[end] yield from trace_path(start, end, prev) yield step def solve(max_turns, field, time): i0, j0 = find_start_pos(field) p0 = i0, j0, 0, 0 prev = {} que = deque([p0]) directions = list(itertools.product((-1, 0, 1), (-1, 0, 1))) while que: p = i, j, vi, vj = que.popleft() for dvi, dvj in directions: vi1, vj1 = vi + dvi, vj + dvj i1, j1 = i + vi1, j + vj1 if not can_go(field, i1, j1): continue p1 = i1, j1, vi1, vj1 if p1 in prev: continue que.append(p1) prev[p1] = p, (dvi, dvj) if field.grids[i1][j1] == Grid.END: return trace_path(p0, p1, prev) return [] def main(): for dvy, dvx in solve(int(input()), read_field(), float(input())): print(dvx, dvy) main()  # ＃ Results ± % time ruby controller.rb benchmark.txt python ../mybfs.py !9349 ["benchmark.txt", "python", "../mybfs.py"] Running 'python ../mybfs.py' against benchmark.txt No. Size Target Score Details ------------------------------------------------------------------------------------- 1 37 x 1 36 0.22222 Racer reached goal at ( 36, 0) in 8 turns. 2 38 x 1 37 0.24324 Racer reached goal at ( 37, 0) in 9 turns. 3 33 x 1 32 0.25000 Racer reached goal at ( 32, 0) in 8 turns. 4 10 x 10 10 0.40000 Racer reached goal at ( 7, 7) in 4 turns. 5 9 x 6 8 0.37500 Racer reached goal at ( 6, 0) in 3 turns. 6 15 x 7 16 0.37500 Racer reached goal at ( 12, 4) in 6 turns. 7 17 x 8 16 0.31250 Racer reached goal at ( 14, 0) in 5 turns. 8 19 x 13 18 0.27778 Racer reached goal at ( 0, 11) in 5 turns. 9 60 x 10 107 0.14953 Racer reached goal at ( 0, 6) in 16 turns. 10 31 x 31 106 0.23585 Racer reached goal at ( 27, 0) in 25 turns. 11 31 x 31 106 0.24528 Racer reached goal at ( 15, 15) in 26 turns. 12 50 x 20 50 0.24000 Racer reached goal at ( 49, 10) in 12 turns. 13 100 x 100 2600 0.01385 Racer reached goal at ( 50, 0) in 36 turns. 14 79 x 63 242 0.24380 Racer reached goal at ( 3, 42) in 59 turns. 15 26 x 1 25 0.32000 Racer reached goal at ( 25, 0) in 8 turns. 16 17 x 1 19 0.52632 Racer reached goal at ( 16, 0) in 10 turns. 17 50 x 1 55 0.34545 Racer reached goal at ( 23, 0) in 19 turns. 18 10 x 7 23 0.34783 Racer reached goal at ( 1, 3) in 8 turns. 19 55 x 55 45 0.17778 Racer reached goal at ( 50, 26) in 8 turns. 20 101 x 100 100 0.14000 Racer reached goal at ( 99, 99) in 14 turns. 21 100000 x 1 1 1.00000 Racer reached goal at ( 0, 0) in 1 turns. 22 50 x 50 200 0.05500 Racer reached goal at ( 47, 46) in 11 turns. ------------------------------------------------------------------------------------- TOTAL SCORE: 6.49643 ruby controller.rb benchmark.txt python ../mybfs.py 3.06s user 0.06s system 99% cpu 3.146 total  • Yes, according to user2357112's comment there's a bug in nneonneo's cycle prevention. As far as I know, the speed is limited by O(√n) which would make your implementation O(n³) on square grids (same as the others, I suppose). I will add a tie breaker to score your submission versus user2357112's later today. – Martin Ender Jul 1 '14 at 8:10 • Btw, are you planning to add another test case? – Martin Ender Jul 1 '14 at 9:19 • @m.buettner No, I dont have a good enough comprehension for this game. So my testcase wont be a interesting one. – Ray Jul 1 '14 at 9:23 • Unfortunately, I had to add another rule. Memory seems to be more limiting than time in this challenge, so I had to set a hard to limit on memory consumption. Any run in which your racer uses more than 1GB of memory will be aborted to the same effect as exceed the time limit. With this rule, your submission is the first to exceed that limit on a tie-breaker of size n=270, which is why you're now behind the other two "optimal" submissions. That being said, your submission is also the slowest of the three, so would have been third anyway, just with a larger tie-breaker. – Martin Ender Jul 1 '14 at 14:45 • Please let me know if you apply any optimisations, so I can rerun the tests. – Martin Ender Jul 1 '14 at 14:45 ## RandomRacer, ~40.0 (averaged over 10 runs) It's not that this bot never finishes a track, but definitely much less often than once in 10 attempts. (I get a non-worst-case score every 20 to 30 simulations or so.) This is mostly to act as a baseline case and to demonstrate a possible (Ruby) implementation for a racer: # Parse initial input target = gets.to_i size = gets.split.map(&:to_i) track = [] size[1].times do track.push gets end time_budget = gets.to_f # Find start position start_y = track.find_index { |row| row['S'] } start_x = track[start_y].index 'S' position = [start_x, start_y] velocity = [0, 0] while true x = rand(3) - 1 y = rand(3) - 1 puts [x,y].join ' '$stdout.flush

first_line = gets
break if !first_line || first_line.chomp.empty?

position = first_line.split.map(&:to_i)
velocity = gets.split.map(&:to_i)
time_budget = gets.to_f
end


Run it with

ruby controller.rb benchmark.txt ruby randomracer.rb


Random Racer 2.0, ~31

Well this isn't going to beat the posted optimal solver, but it is a slight improvement on a random racer. Main difference is that this racer only will consider randomly going where there isn't a wall, unless it runs out of valid places to move, and if it can move to a goal that turn, it will. It also will not move in order to stay in the same spot, unless there is no other move available (unlikely, but possible).

Implemented in Java, compiled with java8, but Java 6 should be fine. No command line parameters. There's a pretty good clusterfuck of hierarchy, so I think I'm doing java right.

import java.util.Scanner;
import java.util.Random;
import java.util.ArrayList;
import java.io.ByteArrayOutputStream;
import java.io.PrintStream;

public class VectorRacing   {
private static Scanner in = new Scanner(System.in);
private static Random rand = new Random();
private static Track track;
private static Racer racer;
private static int target;
private static double time;
public static void main(String[] args)  {
init();
main_loop();
}
private static void main_loop() {
Scanner linescan;
String line;
int count = 0,
x, y, u, v;

while(!racer.lost() && !racer.won() && count < target)  {
Direction d = racer.think();
racer.move(d);
count++;
System.out.println(d);

line = in.nextLine();
if(line.equals("")) {
break;
}
linescan = new Scanner(line);
x = linescan.nextInt();
y = linescan.nextInt();
linescan = new Scanner(in.nextLine());
u = linescan.nextInt();
v = linescan.nextInt();
time = Double.parseDouble(in.nextLine());

assert x == racer.location.x;
assert y == racer.location.y;
assert u == racer.direction.x;
assert v == racer.direction.y;
}
}
private static void init()  {
target = Integer.parseInt(in.nextLine());
int width = in.nextInt();
int height = Integer.parseInt(in.nextLine().trim());
String[] ascii = new String[height];
for(int i = 0; i < height; i++) {
ascii[i] = in.nextLine();
}
time = Double.parseDouble(in.nextLine());
track = new Track(width, height, ascii);
for(int y = 0; y < ascii.length; y++)   {
int x = ascii[y].indexOf("S");
if( x != -1)    {
racer = new RandomRacer(track, new Location(x, y));
break;
}
}
}

public static class RandomRacer extends Racer   {
public RandomRacer(Track t, Location l) {
super(t, l);
}
public Direction think()    {
ArrayList<Pair<Location, Direction> > possible = this.getLocationsCanMoveTo();
if(possible.size() == 0)    {
return Direction.NONE;
}
Pair<Location, Direction> ret = null;
do  {
ret = possible.get(rand.nextInt(possible.size()));
}   while(possible.size() != 1 && ret.a.equals(this.location));
return ret.b;
}
}

// Base things
public enum Direction   {
NORTH("0 -1"), SOUTH("0 1"), EAST("1 0"), WEST("-1 0"), NONE("0 0"),
NORTH_EAST("1 -1"), NORTH_WEST("-1 -1"), SOUTH_EAST("1 1"), SOUTH_WEST("-1 1");

private final String d;
private Direction(String d) {this.d = d;}
public String toString()    {return d;}
}
public enum Cell    {

private final char c;
private Cell(char c)    {this.c = c;}
public String toString()    {return "" + c;}
}

public static class Track   {
private Cell[][] track;
private int target;
private double time;
public Track(int width, int height, String[] ascii) {
this.track = new Cell[width][height];
for(int y = 0; y < height; y++) {
for(int x = 0; x < width; x++)  {
switch(ascii[y].charAt(x))  {
case '#':   this.track[x][y] = Cell.WALL; break;
case '*':   this.track[x][y] = Cell.GOAL; break;
case '.':
case 'S':   this.track[x][y] = Cell.ROAD; break;
default:    System.exit(-1);
}
}
}
}
public Cell atLocation(Location loc)    {
if(loc.x < 0 || loc.x >= track.length || loc.y < 0 || loc.y >= track[0].length) return Cell.OUT_OF_BOUNDS;
return track[loc.x][loc.y];
}

public String toString()    {
ByteArrayOutputStream bos = new ByteArrayOutputStream();
PrintStream ps = new PrintStream(bos);
for(int y = 0; y < track[0].length; y++)    {
for(int x = 0; x < track.length; x++)   {
ps.append(track[x][y].toString());
}
ps.append('\n');
}
String ret = bos.toString();
ps.close();
return ret;
}
}

public static abstract class Racer  {
protected Velocity tdir;
protected Location tloc;
protected Track track;
public Velocity direction;
public Location location;

public Racer(Track track, Location start)   {
this.track = track;
direction = new Velocity(0, 0);
location = start;
}
public boolean canMove() throws GoHereDammitException {return canMove(Direction.NONE);}
public boolean canMove(Direction d) throws GoHereDammitException    {
tdir = new Velocity(direction);
tloc = new Location(location);
tloc.move(tdir);
Cell at = track.atLocation(tloc);
if(at == Cell.GOAL) {
throw new GoHereDammitException();
}
}
public ArrayList<Pair<Location, Direction> > getLocationsCanMoveTo()    {
ArrayList<Pair<Location, Direction> > ret = new ArrayList<Pair<Location, Direction> >(9);
for(Direction d: Direction.values())    {
try {
if(this.canMove(d)) {
}
}   catch(GoHereDammitException e)  {
ret.clear();
return ret;
}
}
return ret;
}
public void move()  {move(Direction.NONE);}
public void move(Direction d)   {
location.move(direction);
}
public boolean won()    {
return track.atLocation(location) == Cell.GOAL;
}
public boolean lost()   {
return track.atLocation(location) == Cell.WALL || track.atLocation(location) == Cell.OUT_OF_BOUNDS;
}
public String toString()    {
return location + ", " + direction;
}
public abstract Direction think();

public class GoHereDammitException extends Exception    {
public GoHereDammitException()  {}
}
}

public static class Location extends Point  {
public Location(int x, int y)   {
super(x, y);
}
public Location(Location l) {
super(l);
}
public void move(Velocity d)    {
this.x += d.x;
this.y += d.y;
}
}

public static class Velocity extends Point  {
public Velocity(int x, int y)   {
super(x, y);
}
public Velocity(Velocity v) {
super(v);
}
if(d == Direction.NONE) return;
if(d == Direction.NORTH || d == Direction.NORTH_EAST || d == Direction.NORTH_WEST)  this.y--;
if(d == Direction.SOUTH || d == Direction.SOUTH_EAST || d == Direction.SOUTH_WEST)  this.y++;
if(d == Direction.EAST || d == Direction.NORTH_EAST || d == Direction.SOUTH_EAST)   this.x++;
if(d == Direction.WEST || d == Direction.NORTH_WEST || d == Direction.SOUTH_WEST)   this.x--;
}
}

public static class Point   {
protected int x, y;
protected Point(int x, int y)   {
this.x = x;
this.y = y;
}
protected Point(Point c)    {
this.x = c.x;
this.y = c.y;
}
public int getX()   {return x;}
public int getY()   {return y;}
public String toString()    {return "(" + x + ", " + y + ")";}
public boolean equals(Point p)  {
return this.x == p.x && this.y == p.y;
}
}

public static class Pair<T, U>  {
public T a;
public U b;
public Pair(T t, U u)   {
a=t;b=u;
}
}
}


The Results (Best case I've seen)

Running 'java VectorRacing' against ruby-runner/benchmark.txt

No.    Size     Target   Score     Details
-------------------------------------------------------------------------------------
1    37 x 1        36   0.38889   Racer reached goal at ( 36, 0) in 14 turns.
2    38 x 1        37   0.54054   Racer reached goal at ( 37, 0) in 20 turns.
3    33 x 1        32   0.62500   Racer reached goal at ( 32, 0) in 20 turns.
4    10 x 10       10   0.40000   Racer reached goal at ( 9, 8) in 4 turns.
5     9 x 6         8   0.75000   Racer reached goal at ( 6, 2) in 6 turns.
6    15 x 7        16   2.00000   Racer did not reach the goal within 16 turns.
7    17 x 8        16   2.00000   Racer hit a wall at position ( 8, 2).
8    19 x 13       18   0.44444   Racer reached goal at ( 16, 2) in 8 turns.
9    60 x 10      107   0.65421   Racer reached goal at ( 0, 6) in 70 turns.
10    31 x 31      106   2.00000   Racer hit a wall at position ( 25, 9).
11    31 x 31      106   2.00000   Racer hit a wall at position ( 8, 1).
12    50 x 20       50   2.00000   Racer hit a wall at position ( 27, 14).
13   100 x 100    2600   2.00000   Racer went out of bounds at position ( 105, 99).
14    79 x 63      242   2.00000   Racer went out of bounds at position (-2, 26).
15    26 x 1        25   0.32000   Racer reached goal at ( 25, 0) in 8 turns.
16    17 x 1        19   2.00000   Racer went out of bounds at position (-2, 0).
17    50 x 1        55   2.00000   Racer went out of bounds at position ( 53, 0).
18    10 x 7        23   2.00000   Racer went out of bounds at position ( 10, 2).
19    55 x 55       45   0.33333   Racer reached goal at ( 4, 49) in 15 turns.
20    50 x 50      200   2.00000   Racer hit a wall at position ( 14, 7).
-------------------------------------------------------------------------------------
TOTAL SCORE:             26.45641

• Yes, I got it running, although I had to run it from the directory where the .class file is for some reason (instead of the directory where the controller is). Ping me (with a comment) if you decide to add a testcase, so I can add it to the benchmark. Your score was about 33 over 10 runs (see leaderboard), but this might well change with every new test track that is added to the benchmark. – Martin Ender Jun 27 '14 at 14:11
• Ah got it to run from the other directory, too. For those not familiar with Java on the command line: java -cp path/to/class/file VectorRacing – Martin Ender Jun 27 '14 at 14:19
• Ah, yeah I made a ton of classes (13, to be exact). I was always running your script from my classes directory, so I didn't actually test that. I may make a test case, but I think I'll try to make a racer that isn't random first. – pseudonym117 Jun 27 '14 at 14:24
• Sure. If you do, please add it as a separate answer. (And feel free to add one test case with each of them.) – Martin Ender Jun 27 '14 at 14:27
• Unfortunately, I had to add another rule. Memory seems to be more limiting than time in this challenge, so I had to set a hard to limit on memory consumption. Any run in which your racer uses more than 1GB of memory will be aborted to the same effect as exceed the time limit. For the current set of tracks, your score has not been affected by this change (and likely won't ever be). – Martin Ender Jul 1 '14 at 14:42